Water quality parameters
Water quality is described by physical, chemical and biological parameters. Each tells a different part of the story, and which ones matter depends on whether the water is for drinking, irrigation, industry, recreation or aquatic life, or is wastewater being discharged.
Physical parameters include temperature, which affects oxygen solubility and reaction rates; turbidity, a measure of cloudiness from suspended particles; colour, taste and odour; and solids, divided into suspended and dissolved, and into volatile (organic) and fixed (mineral) fractions. Electrical conductivity is a quick proxy for dissolved salts.
Chemical parameters include pH; alkalinity, the capacity to neutralise acid, important for coagulation and corrosion; hardness from calcium and magnesium; dissolved oxygen, vital for aquatic life; BOD and COD for organic pollution; nutrients, nitrogen and phosphorus in their various forms, which cause algal blooms and eutrophication; chloride, sulphate, fluoride, iron and manganese; and toxic substances such as heavy metals, pesticides and emerging contaminants like pharmaceuticals and microplastics. Biological parameters include indicator bacteria (total and faecal coliforms, E. coli), algae and, for rivers, indices based on the diversity of aquatic insects.
Monitoring programmes choose parameters, locations, frequency and methods to meet a clear objective, follow standard sampling and preservation procedures, and use accredited laboratories. Indian surface water bodies are classified by designated best use, each with criteria for parameters such as dissolved oxygen, BOD and coliforms. Data is only useful if it is consistent over time and openly reported.
Preserve and chill samples as the method requires and test BOD and bacteria within the specified holding time.
Total solids = suspended + dissolvedTDS (mg/L) ≈ 0.55 to 0.7 × EC (µS/cm)DO saturation falls as temperature risesUnionised ammonia fraction rises with pH and temperature
Comparing results from different laboratories without checking that they used the same methods and units.
Problem. A river sample has EC = 820 µS/cm. Estimate TDS with a factor of 0.64. If DO is 5.1 mg/L and saturation at the water temperature is 8.0 mg/L, find the percent saturation and oxygen deficit.
- TDS ≈ 0.64 × 820 = 525 mg/L.
- Percent saturation = 5.1 / 8.0 × 100 = 64 percent.
- Oxygen deficit = 8.0 − 5.1 = 2.9 mg/L.
- This suggests moderate organic pollution; check BOD and coliform results.
Answer. TDS ≈ 525 mg/L; DO 64 percent of saturation; deficit 2.9 mg/L
Quick quiz
1What does conductivity indicate?
The amount of dissolved salts.
2Which nutrients cause eutrophication?
Nitrogen and phosphorus.
3Name one biological indicator of faecal pollution.
E. coli (or faecal coliforms).
- IS 10500 (India): Specification for drinking water quality.
- CPHEEO Manual on Water Supply and Treatment (India): Central government manual for planning, design and operation of public water supply.
Clauses and documents to read
- IS 2296:1982 (tolerance limits for inland surface waters, cited by number)
- IS 3025 (various parts) (methods of test for water and wastewater)
- CPCB designated best use classification (cited by name)
Cited by number and clause only. We do not reproduce clause text, tables or figures (BIS Act 2016, s.10(5) and s.11). Check the current edition and amendments before use.
References and further reading
Standard texts, listed in no order of merit. We do not copy text, figures, tables or solved problems from any book; our explanations and examples are original.
- H. S. Peavy, D. R. Rowe, G. Tchobanoglous, Environmental Engineering, McGraw Hill, 1985.
- M. L. Davis, D. A. Cornwell, Introduction to Environmental Engineering, McGraw Hill, latest edition.
Free and official sources
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